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Transcript of calculaton of u valve thermal brizzing
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Module 3.4
Calculation of U-value
Construction with thermal bridging
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Learning Outcomes
On successful completion of this modulelearners will be able to
Describe the procedure for the calculation of
U-values for constructions with thermal bridging.
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Forward.
This document uses extracts from, and makesreferences to, EN ISO 6946 : 2007
Building components and building elements -Thermal resistance and thermal transmittance -
Calculation method. The content of this document is not a substitute
for the standard. To properly apply the standardone must have the complete document.
Standards available from www.iso.org.
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Introduction to thermal bridging.
Thermal bridging occurs in building envelopes whenmaterials with high thermal conductivity, such as steeland concrete, create pathways for heat loss that bypassthermal insulation.
When these poor insulating materials provide anuninterrupted short circuit between the interior andexterior of a building, the thermal bridge can result in
a) accelerated heat loss through that area,
b) localised cold spots on the interior of a wall that maycause a risk of condensation.
This effect is most significant in cold climates during thewinter when the indoor-outdoor temperature differenceis greatest. Reference: http://wiki.aia.org/WikiPages/ThermalBridging.aspx
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- continued.
Thermal bridging can also occurs in buildingenvelopes when gaps or breaks in theinsulation envelope create pathways for heatloss that bypass thermal insulation.
The reduced or missing insulation can allowwarm inside air make contact with cold externalsurfaces.
This bypass of the insulation will also causeaccelerated heat loss through that area, andlocalised cold spots on the interior.
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Identification of thermal bridging.
Take any heat loss surface, i.e. floor, wall, roof, etc. Thermal bridging must be considered if there is more
than one possible heat flow path through that surface.
Typical construction details around doors and windows
are treated separately. The are not included whencalculating thermal bridging for that wall.
If the thermal bridge allows for two or more possibleheat flow paths, in the main body of the floor, wall, roof.etc. then its effect must be considered.
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continued.
Take heat loss through the timber frame constructionshow on the next page.
One resistance layer is made up of two very differentmaterials, timber and mineral wool insulation.
The poor insulator, timber, provides an uninterruptedshort circuit through the insulation, between theinterior and exterior of a building.
The timber create pathways for heat loss that bypassthe insulation.
This reduces the effectiveness of the insulation and itseffect must be included in the calculation of the U-valueof the wall.
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Procedure for Calculation of U-value Thermal bridging
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- Procedure continued.
For a thermally bridged construction, the
U-value is calculated following five steps.Reference Examples of U-value calculations using BS EN ISO 6946:1997.
Step 1 Calculate Upper resistance limit(R
upper)
Step 2 Calculate Lower resistance limit (Rlower
)
Step 3 Calculate Total thermal resistance (RT)
Step 4 Calculate corrections / adjustments for
a) air gaps penetrating the insulation.
b) mechanical fixings penetrating the insulation.
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- Procedure continued. Step 1.
Calculate the upper resistance limit (Rupper
)
This is done by combining, in parallel, the totalresistances of all possible heat flow paths
(i.e. sections) through the sample construction onthe previous pages.
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- Procedure continued.
Step 2.
Calculate the lower resistance limit (Rlower
)
This is done by combining, in parallel, theresistances of the heat flow paths of each layerseparately and then summing the resistance of alllayers (i.e. sections) through the givenconstruction.
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- Procedure continued.
Step 3.
Calculate the total thermal resistance (RT)
This is done using
RT
= Rupper
+ Rlower
2
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- Procedure continued.
Step 4Procedure - Calculate, where appropriate,
a) corrections for air gaps ( Ug ) in the insulation
layer.b) correction for mechanical fasteners ( Uf ) in theinsulation layer.
Note: Procedures for corrections may vary between
countries. One procedure is described in BRE442 Conventions for U-value calculations, 2006, sections 4.9.1,4.9.2 and 4.9.3.
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Procedure continued.
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Procedure continued.
BRE443 section 4.9.2 and 4.9.3 suggests that the effectof wall ties is negligible in an un-insulated cavity, and inany cavity if plastic ties are used.
Otherwise the effect of wall ties needs to be considered.
The correction requires knowledge of the thermalconductivity of the ties, their cross-sectional area, andthe number per square metre of wall.
For detailed procedure see BRE442 Conventions for
U-value calculations, 2006, sections 4.9.2 and 4.9.3.
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- Procedure continued.
Step 5
Calculate the final U-value from,
U = ( 1 / RT
) + Ug
+ Uf
Note: Ug
and Ufcan be omitted,
if taken together,
they amount to less than 3% of the U-value found from U =1 / R
T
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Example - Step 1 Calculating Rupper
Procedure.
Combining, in parallel, the total resistances of allpossible heat flow paths (i.e. sections) through the
given timber frame construction For this construction shown on the preceding
sketch, thermal bridging occurs only at onelayer, that is the layer made up partly of 38 x
140mm timber and partly by 140mm mineralwool.
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Example Step 1 - Rupper
continued.
For this construction there are two possible heatflow paths, between inside and outside.
First heat flow path is through the insulation, i.e.
2 x 12.5mm plasterboard+ 140mm mineral wool insulation
+ 19mm plywood
+ 50mm slightly ventilated air cavity+ 102mm brick outer leaf.
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Example Step 1 - Rupper
continued.
Second heat flow path is through the timber, i.e.
2 x 12.5mm plasterboard
+ 38 x 140mm timber studs
+ 19mm plywood
+ 50mm slightly ventilated air cavity
+ 102mm brick outer leaf.
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Example Step 1 - Rupper
continued.
The two parallel heat flow paths are shownconceptually below.
Since the relative cross sectional area of the timber isnot the same as that of the mineral wool insulation, thefractional areas of each must be considered whencalculating R
upper.
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Example Step 1 - Rupper
continued.
The upper limit of resistance is calculated from
Rupper
= 1
F1
+ F2
R1
R2
F1
= the fractional area of first (insulation) heat loss path
F2 = the fractional area of second (timber) heat loss path
R1
= the total resistance of first (insulation) heat loss path
R2
= the total resistance of second (timber) heat loss path
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Example Step 1 - Rupper
continued.
Fractional area of timber and mineral wool.
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Example Step 1 - Rupper
continued.
When considering the fractional area of timberdifferent countries may use differentprocedures.
To provide inputs for this worked example theprocedures described in BRE443 section 4.5.1have been used.
The wall area is taken as a number of
repeatable rectangles 2400mm high x 400mmwide.
The percentage timber in this repeatable area iscalculated as follows.
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Example Step 1 - Rupper
continued.
Fractional area of timber F2 and mineral wool F1.
Overall repeatable area
2.400 x 0.400 = 0.960 m2
Area of timber
Vertical 2.400 x 0.038 = 0.091 m2
Horizontal 0.362 x 0.038 x 4 = 0.055 m2
Total 0.146 m2
Fractional area of timber (F2) (0.146 / 0.960) 0.15 Fractional area insulation(F
1) ( 1 0.15 ) 0.85
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Example Step 1 - Calculation of R1
- Sum of
thermal resistance through insulation.
Layer / Surface
Thickness (d)
( m )
Conductivity ()
( W / m K )
Resistance (R=d/)
( m2 K / W )
External surface (Rse) - - - - - - 0.040
Outer leaf brick 0.102 0.77 0.132
Air cavity (Ra) 0.500 - - - 0.090
Plywood 0.019 0.13 0.146
Insulation 0.140 0.038 3.684
Plaster board 0.025 0.25 0.100
Internal surface (Rsi) - - - - - - 0.130
Total Resistance - - - - - - 4.322
Note: Values of Rse, Ra and Rsi taken from I.S. EN ISO 6946 Table 1 and Table 2.Thermal conductivity values used are indicative only.
Certified values should be used, if available.
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Example Step 1 - Calculation of R2
- Sum of
thermal resistance through timber.
Layer / Surface
Thickness (d)
( m )
Conductivity ()
( W / m K )
Resistance (R=d/)
( m2 K / W )
External surface (Rse) - - - - - - 0.040
Outer leaf brick 0.102 0.77 0.132
Air cavity (Ra) 0.500 - - - 0.090
Plywood 0.019 0.13 0.146
Timber 0.140 0.13 1.077
Plaster board 0.025 0.25 0.100
Internal surface (Rsi) - - - - - - 0.130
Total Resistance - - - - - - 2.525
Note: Values of Rse, Ra and Rsi taken from I.S. EN ISO 6946 Table 1 and Table 2.Thermal conductivity values used are indicative only.
Certified values should be used, if available.
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Example Step 1 - Calculation of Rupper
The upper limit of resistance is now calculated fromR
upper= 1
Finsul
+ Ftimber
Rinsul Rtimber
Rupper
= 1
0.85 + 0.15
4.322 2.525
Rupper
= 3.906 m2 K / W
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Example Step 2 Calculating Rlower
Procedure.
Combining, in parallel, the resistances of the heatflow paths of each layer separately and then sum
the resistance of all layers (i.e. sections) throughthe given construction.
For this construction, thermal bridging occursonly at one layer, that is the layer made up
partly of 38 x 140mm timber and partly by140mm mineral wool.
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Example Step 2 Calculating R
lower continued
For this construction there is only one layer withtwo possible heat flow paths, i.e. the layer witha combination of 140mm mineral wool
insulation and 38 x 140 timber studs. The heat flow path will be
2 x 12.5mm plasterboard
+ (Combination of mineral wool and timber)
+ 19mm plywood
+ 50mm slightly ventilated air cavity
+ 102mm brick outer leaf.
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Example Step 2- Calculating R
lower continued.
The heat flow path for the timber frameexample is shown conceptually below.
Since the relative cross sectional area of the timber isnot the same as that of the mineral wool insulation, thefractional areas of each must be considered whencalculating R
lower. This has already been calculated
when calculating Rupper
on previous slides.
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Example Step 2- Calculating R
lower continued.
The resistance of the bridged layer is now calculatedfrom
Rbridged
= 1
Finsul + Ftimber
Rinsul
Rtimber
F insul = the fractional area of first (insulation) heat loss pathF
timber= the fractional area of second (timber) heat loss path
R insul = the total resistance of first (insulation) heat loss pathR timber = the total resistance of second (timber) heat loss path
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Example Step 2- Calculating R
lower continued
From previous timber frame sketch andcalculations in this module.
Fractional area of timber 0.15Fractional area of insulation 0.85
Resistance of insulation 3.684 m2K/W
Resistance of timber 1.077 m2
K/W
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Example Step 2 - Rlower
continued.
The resistance of the bridged layer is now calculatedfrom
Rbridged
= 1
Finsul
+ Ftimber
Rinsul RtimberR
bridged= 1
0.85 + 0.15
3.684 1.077
Rbridged
= 2.703 m2 K / W
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Example Step 2 - Rlower
continued.
Layer / Surface
Thickness (d)
( m )
Conductivity ()
( W / m K )
Resistance (R=d/)
( m2 K / W )
External surface (Rse) - - - - - - 0.040
Outer leaf brick 0.102 0.77 0.132Air cavity (Ra) 0.500 - - - 0.090
Plywood 0.019 0.13 0.146
Bridged layer - - - - - - 2.703
Plaster board 0.025 0.25 0.100
Internal surface (Rsi) - - - - - - 0.130
Total Resistance - - - - - - 3.341
Sum of thermal resistance through all layers
Rlower= 3.341 m2
K / W
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Step 3 Calculate total thermal resistance.
Procedure.Calculate the total thermal resistance (R
T) using
RT
= Rupper
+ Rlower
2
RT
= 3.906 + 3.341
2
Total thermal resistance RT
= 3.624 m2 K / W
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Step 4 Calculate corrections.
Procedure - Calculate, where appropriate,
a) corrections for air gaps ( Ug
) in the insulation
layer. That is if a gaps exceeding
5mm width penetrates the insulation.
b) correction for mechanical fasteners ( Uf) in the
insulation layer.
Note: Procedures for corrections may vary betweencountries. One procedure is described in BRE442 Conventions for U-value calculations, 2006, sections 4.9.1,4.9.2 and 4.9.3.
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Step 4 - Corrections - continued.
Corrections for air gaps ( Ug
) in the insulation
layer.
In this example no corrections are applicable, i.
e. the insulation is installed in such a way thatthere are no air gaps penetrating the entireinsulation layer and no air circulation is possibleon the warm side of the insulation.
For the three possible levels of correction for air gaps see I.S. EN ISO 6946 AnnexD.
For examples of such corrections see
Examples of U-value calculations using BS EN ISO 6946 : 1997 published by BRE.
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Step 4 - Corrections - continued.
Correction for mechanical fasteners ( Uf) in
the insulation layer.
In this example no corrections are applicable, i.
e. no mechanical fasteners penetrating theentire insulation layer.
For the formula related correction for mechanical fastners see I.S EN ISO 6946Annex D.
For examples of such corrections see
Examples of U-value calculations using BS EN ISO 6946 : 1997 published by BRE.
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Step 5 Calculate the final U-value
Procedure
U = ( 1 / RT
) + Ug
+ Uf
U = ( 1 / 3.624 ) + 0 + 0
U = 0.28 W /m2 K
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Module Summary.
When calculating U-values for constructionswith thermally bridged elements, 5 steps areused.
Task.Outline each of these 5 steps.
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Module Summary.
Step 1 calculates the upper resistance limitR
upper.
Task.
Describe this step in detail and draw a sketch toshow how the resistance paths should bearranged.
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Module Summary.
Step 2 calculates the upper resistance limitR
lower.
Task.
Describe this step in detail and draw a sketch toshow how the resistance paths should bearranged.
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Module Summary.
Step 3 calculates the total thermal resistancelimit R
T.
Task.
Describe the formula needed to calculate RT.
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Module Summary.
Step 4 calculates the corrections for
a) air gaps penetrating the insulation
b) mechanical fixings penetrating the insulation.
Task.Describe some examples where these correctionfactors need not be applied.
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Module Summary.
Step 5 calculates the thermal transmittancevalue of U-value.
Task.
Describe the formula needed to calculate theU-value.
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National procedures UK, Ireland.
When considering the fractional area of timberincluded in calculations BRE 443 Section 4.5.1states :-
For timber frame construction which conforms with
the designs in Accredited Construction Details,additional heat loss at corners, window surrounds,between floors etc is limited and the associatedtimbers are not counted as part of the timber
fraction. This is comparable with masonryconstruction where items such as lintels and cavityclosers are not counted.
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- continued.
BRE 443 suggests a default fraction for typicaltimber frame construction as 0.15.
A lower fraction of 0.125 may be used if specific
conditions are met. For further information see BRE 443 section
4.5.
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Acknowledgements:
The authors and publishers of this documentwish to thank the National Standards Authorityof Ireland for permission to reproduce extracts
from copyright materialEN ISO 6946 : 2007Building components and building elements -Thermal resistance and thermal transmittance -Calculation method.
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References:
International standards.
I.S. EN ISO 6946 : 2007 Building components andbuilding elements - Thermal resistance and thermaltransmittance - Calculation method.
National standards + others.
BRE 443 : 2006 Convention for U-valuecalculations. ISBN 1 86081 924 9
Examples of U-value calculations using BS ENISO 6946 : 1997 published by BRE.
http://wiki.aia.org/WikiPages/ThermalBridging.aspx